0
$\begingroup$

I have been trying to understand how to solve the exercise. I took a look at the answer and apparently you are supposed to solve $y(t) = 0$.

I do not understand this at all, I tried watching some videos on parametric equations but it just doesn‘t seem to correlate.

I mean, yes in a normal function I would understand if we were to equal the function to zero and plug the resulting $x$ into the original equation to find the y-coordinates. But why would we only search for $y(t)=0$ here and not $x(t)=0$

I would appreciate some tips on how to solve these, thanks in advance!

Exercise

$\endgroup$
  • $\begingroup$ You are correct in saying $y(t)=0$. Now you just need to plug in $y(t)= \sin{2t} - 2\cos{t} = 0$ and find a way to calculate the value of the expression $\cos{2t} - 3\sin{t} - 1$. This can be done in many ways, for example replacing $\sin{2t}$ by $2\sin{x}\cos{x}$. $\endgroup$ – Matti P. May 15 '18 at 5:16
  • $\begingroup$ Hey Matti, thanks but well I don‘t understand why we would take y(t)=0 and not say x(t)=0. Like I said I don‘t really get it. Edited the question with further elaboration..... $\endgroup$ – user549904 May 15 '18 at 5:19
  • $\begingroup$ @user549904 because the question says "the curve $K$ shares two points $A$ and $B$ with the $x$-axis." When does the curve hit the $x$-axis? Precisely when the curve's $y$ values are exactly zero, i.e. $y(t)=0$. $\endgroup$ – Pixel May 15 '18 at 5:26
  • $\begingroup$ It's important to understand what are the equations of the $x$-axis and the $y$-axis. The equation of the $x$-axis is $y=0$ and the equation of the $y$-axis is $x=0$. If you just look at the picture, it should become clear. $\endgroup$ – Matti P. May 15 '18 at 5:31
1
$\begingroup$

You have already been told $t=\frac{\pi}{2}$ is for $A$ so you just need to evaluate $x$ and $y$ there.

$$ x( \frac{\pi}{2} ) = \cos \pi - 3 \sin \frac{\pi}{2} - 1 = -1 - 3 -1 = -5\\ y( \frac{\pi}{2} ) = \sin \pi - 2 \cos \frac{\pi}{2} = 0 - 0 = 0 $$

So yes $A$ does intersect the $x$-axis because $y=0$ there. The x-axis has the equation $y=0$.

What is the other time for which $y(t)=0$

$$ y (t) = \sin 2 t - 2 \cos t $$

What about $\frac{3\pi}{2}$? It is similar to $\frac{\pi}{2}$ but on the other side.

$$ y (\frac{3\pi}{2}) = \sin (3 \pi) - 2 \cos \frac{3\pi}{2} = 0 - 0 $$

Good evaluate $x ( \frac{3\pi}{2}) = -1 - 3 (-1) - 1 = 1$

$\endgroup$
  • $\begingroup$ Okay that makes sense! Finally something I can understand. Thanks Husain! $\endgroup$ – user549904 May 15 '18 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.